标签:queue mat log clu vector 打开 roo 数据 getc
[BZOJ1095][ZJOI2007]Hide 捉迷藏
试题描述
输入
输出
输入示例
8 1 2 2 3 3 4 3 5 3 6 6 7 6 8 7 G C 1 G C 2 G C 1 G
输出示例
4 3 3 4
数据规模及约定
对于100%的数据, N ≤100000, M ≤500000。
题解
这题一看就肯定是动态点分治。
然而 sb 的我并没有自己想出具体做法。。。
建立好重心树后,我们维护三种堆:(以下的描述都是在重心树上的,与原树无关)
1. 每个节点一个堆维护子树中所有节点到它父亲的距离;
2. 每个节点一个堆维护所有儿子的子树到自己的最大距离(维护的是这些最大距离的集合);
3. 一个堆维护每个节点对应的子树中经过它的最长链(维护的是这些最长链的集合)。
#include <iostream> #include <cstdio> #include <algorithm> #include <cmath> #include <stack> #include <vector> #include <queue> #include <cstring> #include <string> #include <map> #include <set> using namespace std; const int BufferSize = 1 << 16; char buffer[BufferSize], *Head, *Tail; inline char Getchar() { if(Head == Tail) { int l = fread(buffer, 1, BufferSize, stdin); Tail = (Head = buffer) + l; } return *Head++; } int read() { int x = 0, f = 1; char c = Getchar(); while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = Getchar(); } while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = Getchar(); } return x * f; } #define maxn 100010 #define maxm 200010 #define maxlog 18 int n, m, head[maxn], nxt[maxm], to[maxm]; void AddEdge(int a, int b) { to[++m] = b; nxt[m] = head[a]; head[a] = m; swap(a, b); to[++m] = b; nxt[m] = head[a]; head[a] = m; return ; } int mnd[maxlog][maxn<<1], clo, dfn[maxn], dep[maxn], Log[maxn<<1]; void build(int u, int pa) { dfn[u] = ++clo; mnd[0][clo] = dep[u]; for(int e = head[u]; e; e = nxt[e]) if(to[e] != pa) dep[to[e]] = dep[u] + 1, build(to[e], u), mnd[0][++clo] = dep[u]; return ; } void rmq_init() { Log[1] = 0; for(int i = 2; i <= clo; i++) Log[i] = Log[i>>1] + 1; for(int j = 1; (1 << j) <= clo; j++) for(int i = 1; i + (1 << j) - 1 <= clo; i++) mnd[j][i] = min(mnd[j-1][i], mnd[j-1][i+(1<<j-1)]); return ; } int cdist(int a, int b) { int l = dfn[a], r = dfn[b]; if(l > r) swap(l, r); int t = Log[r-l+1]; return dep[a] + dep[b] - (min(mnd[t][l], mnd[t][r-(1<<t)+1]) << 1); } int rt, size, f[maxn], siz[maxn]; bool vis[maxn]; void getroot(int u, int pa) { siz[u] = 1; f[u] = 0; for(int e = head[u]; e; e = nxt[e]) if(to[e] != pa && !vis[to[e]]) { getroot(to[e], u); siz[u] += siz[to[e]]; f[u] = max(f[u], siz[to[e]]); } f[u] = max(f[u], size - siz[u]); if(f[rt] > f[u]) rt = u; return ; } void dfs(int u, int pa) { siz[u] = 1; for(int e = head[u]; e; e = nxt[e]) if(to[e] != pa && !vis[to[e]]) dfs(to[e], u), siz[u] += siz[to[e]]; return ; } int fa[maxn]; void solve(int u) { vis[u] = 1; for(int e = head[u]; e; e = nxt[e]) if(!vis[to[e]]) { dfs(to[e], u); f[rt = 0] = size = siz[to[e]]; getroot(to[e], u); fa[rt] = u; solve(rt); } return ; } priority_queue <int> tofa[maxn], tofa_del[maxn], son[maxn], son_del[maxn], ans, ans_del; bool has[maxn]; void tofadel(int u, int d) { tofa_del[u].push(d); while(!tofa[u].empty() && !tofa_del[u].empty() && tofa[u].top() == tofa_del[u].top()) tofa[u].pop(), tofa_del[u].pop(); return ; } void sondel(int u, int d) { son_del[u].push(d); while(!son[u].empty() && !son_del[u].empty() && son[u].top() == son_del[u].top()) son[u].pop(), son_del[u].pop(); return ; } void ansdel(int d) { ans_del.push(d); while(!ans.empty() && !ans_del.empty() && ans.top() == ans_del.top()) ans.pop(), ans_del.pop(); return ; } int upans(int u) { if(son[u].empty()) return -1; int t1 = son[u].top(); son[u].pop(); while(!son[u].empty() && !son_del[u].empty() && son[u].top() == son_del[u].top()) son[u].pop(), son_del[u].pop(); if(son[u].empty()){ son[u].push(t1); return -1; } int t2 = son[u].top(); son[u].push(t1); return t1 + t2; } void update(int s) { int tmp = upans(s); if(tmp >= 0) ansdel(tmp); if(!has[s]) son[s].push(0); else sondel(s, 0); tmp = upans(s); if(tmp >= 0) ans.push(tmp); for(int u = s; fa[u]; u = fa[u]) { int d = cdist(s, fa[u]); tmp = upans(fa[u]); if(tmp >= 0) ansdel(tmp); if(!has[s]) { if(tofa[u].empty()) son[fa[u]].push(d); else if(tofa[u].top() < d) sondel(fa[u], tofa[u].top()), son[fa[u]].push(d); tofa[u].push(d); } else { tofadel(u, d); if(tofa[u].empty()) sondel(fa[u], d); else if(tofa[u].top() < d) sondel(fa[u], d), son[fa[u]].push(tofa[u].top()); } tmp = upans(fa[u]); if(tmp >= 0) ans.push(tmp); } has[s] ^= 1; return ; } int main() { n = read(); for(int i = 1; i < n; i++) { int a = read(), b = read(); AddEdge(a, b); } build(1, 0); rmq_init(); f[rt = 0] = size = n; getroot(1, 0); solve(rt); for(int i = 1; i <= n; i++) update(i); int q = read(); while(q--) { char cmd = Getchar(); while(!isalpha(cmd)) cmd = Getchar(); if(cmd == ‘G‘) printf("%d\n", ans.empty() ? -1 : ans.top()); else update(read()); } return 0; }
1A 了好爽 2333333
标签:queue mat log clu vector 打开 roo 数据 getc
原文地址:http://www.cnblogs.com/xiao-ju-ruo-xjr/p/6663759.html